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I am reading the Exterior Measure section by Stein and Shakarchi (2009).

In their definition an exterior of measure of any subset $E$ of $\mathbb{R}^d$ is

$$m_*(E)=\inf\sum^\infty_{j=1}|Q_j|.$$ where the infimum is taken over all countable coverings $E\subset\bigcup^\infty_{j=1}Q_j$.

My Question: I want to show $|Q|\leq m_*(Q) $. |.| is the volume of a cube. Why is it enough to show $|Q|\leq\sum^\infty_{j=1}|Q_j|$ for an aribtrary covering $Q\subset\bigcup_jQ_j$?

Isn't the following true, so where does sufficiency come from? What am I missing?

$$|Q|\leq\sum^\infty_j|Q_j|\leq\inf\sum^\infty_j|Q_j|.$$

Reference: $\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

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  • $\begingroup$ Unless I misunderstand your question, it's just the definition of infimum. $\endgroup$ Commented Sep 7, 2019 at 22:13
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    $\begingroup$ Your last inequality is incorrect. It is not true that $$\sum|Q_j| \leq \inf \sum |Q_j|.$$The opposite inequality is true: $$\inf \sum |Q_j| \leq \sum|Q_j|$$ Note that if $|Q| \leq \sum |Q_j|$ for an arbitrary covering, then $|Q|$ is a lower bound for the set $\{\sum|Q_j| : (Q_j)_{j=1}^{\infty}\text{ a countable covering}\}$, and therefore $|Q|$ cannot be larger than the greatest lower bound (the infimum) of that set. $\endgroup$
    – user169852
    Commented Sep 7, 2019 at 22:21
  • $\begingroup$ @Bungo Thank you for the response! Can I ask a clarification question? We are starting with a subset $E$. So the way I am thinking about the countable covering is that we are looking at a collection of cube-size pillows to cover $E$. If you are looking at the entire collection of such set of pillows (i.e. cover) and asking what is the infimum, what does it mean to take the greatest lower bound of this set? $\endgroup$ Commented Sep 8, 2019 at 21:09
  • $\begingroup$ @Bungo As for the timing, are we taking the infimum from the set of countable covering and then once we find the infimum, we are taking the sumo f the volumes of the cube of THAT infimum? $\endgroup$ Commented Sep 8, 2019 at 21:15
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    $\begingroup$ @FrankSwanton Because this subtlety means that when making arguments / proofs about coverings, you often have to take into account this $\epsilon$ - you can't necessarily find a covering for which $\sum |Q_i| = m^*(E)$, but you can get as close as you like. $\endgroup$
    – user169852
    Commented Sep 8, 2019 at 21:41

4 Answers 4

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Expanding my comments into an answer:

Your last inequality is incorrect. It is not true that $$\sum^\infty_{j=1}|Q_j| \leq \inf\sum^\infty_{j=1}|Q_j|.$$ The opposite inequality is true: if we define $$\mathcal{C}_Q = \left\{\sum^\infty_{j=1}|Q_j|:(Q_j)_{j=1}^{\infty} \text{ a countable covering of $Q$}\right\},$$ then for any covering $(Q_j) \in \mathcal{C}_Q$ we have $$\inf\mathcal{C}_Q \leq \sum^\infty_{j=1}|Q_j|.$$

Once we have shown that $|Q| \leq \sum_{j=1}^{\infty}|Q_j|$ for any countable covering $(Q_j)$ in $\mathcal C_Q$, this will imply that $|Q|$ is a lower bound of $\mathcal{C}_Q$, and therefore $|Q|$ cannot be larger than the greatest lower bound of $\mathcal{C}_Q$, hence $$|Q| \leq \inf \mathcal{C}_Q \leq \sum_{j=1}^{\infty}|Q_j|.$$

Note that $\mathcal{C}_Q$ is a set of nonnegative numbers, not a set of sets. Each number in $\mathcal{C}_Q$ is the total volume $\sum|Q_j|$ of some countable covering of $Q$. Then $\inf \mathcal C_Q$ is the greatest lower bound of this set of numbers. It is not necessarily true that there is a covering $(Q_j)$ that achieves this lower bound. In general, for any covering $(Q_j)$ we will have strict inequality: $$\inf \mathcal C_Q < \sum |Q_j|$$ However, by definition of infimum / greatest lower bound, it is true that for any $\epsilon > 0$, we can find a covering $(Q_j)$ such that $$\inf \mathcal C_Q \leq \sum |Q_j| \leq \inf \mathcal C_Q + \epsilon$$

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  • $\begingroup$ Excellent. Thanks for the response! $\endgroup$ Commented Sep 9, 2019 at 18:45
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It's just the definition of the infimum. If you know that $$|Q|\leq \sum\limits_{j=1}^\infty |Q_j|$$ for any countable covering of cubes, then $$|Q|\leq \inf_{\text{countable coverings} \{Q_j\}}\sum\limits_{j=1}^\infty |Q_j|=m^*(Q).$$ This is because the left-hand side is independent of such coverings. Here's how to see it explicitly:

Remember that something is the infimum if adding any $\epsilon$ to it allows for something smaller. In particular, for any $\epsilon>0,$ there exists a countable covering of $Q$ by sets $\{Q_j\}$ so that $$\sum\limits_{j=1}^\infty |Q_j|<m^*(Q)+\epsilon.$$ Since $$|Q|\leq \sum\limits_{j=1}^\infty |Q_j|,$$ we really have that $$|Q|<m^*(Q)+\epsilon,$$ for all $\epsilon>0.$ Hence, $$|Q|\leq m^*(Q).$$

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  • $\begingroup$ Thank you for the response!! $\endgroup$ Commented Sep 9, 2019 at 18:46
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You are taking $Q$ and you are covering it with a numerable set $\{Q_j\}_{j=1}^{\infty}$. What you are sure is that they cover Q but some $Q_j$ may overlap another one so this is because $$|Q| \le \sum_{j=1}^{\infty} |Q_j|$$

This is still true if you take the inf of that sum, so this is why

$$|Q| \le inf\{\sum_{j=1}^{\infty} |Q_j|\} = m_*(Q)$$

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  • $\begingroup$ Thank you for the response!! $\endgroup$ Commented Sep 9, 2019 at 18:45
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The infimum is the greatest lower bound of a set, by definition.

So if you prove your claim, $|Q|$ is one lower bound of that set and so the infimum, $m_{\ast}(Q)$ by definition, is at least as big, so $|Q| \le m_\ast(Q)$.

Where the set (of (extended) reals) in question is

$$\{\sum_j |Q_j| :\, \{Q_j\} \text{ a countable covering of } Q \text{ by (basic) cubes}\}$$ of course.

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  • $\begingroup$ Thank you for the response!! $\endgroup$ Commented Sep 9, 2019 at 18:46

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